Đáp án: $x=5_{}$
Giải thích các bước giải: $2\sqrt{x-1}+\sqrt{x-4}=x_{}$ $ĐK:_{}$ $x_{}$ $\neq1$ $và_{}$ $x_{}$ $⇔\neq4$ $x_{}$ $\neq0$
$⇔4(x-1)+4\sqrt{(x-1)(x-4)}+x-4=x^2_{}$
$⇔4x-4+4\sqrt{x^2-4x-x+4}+x-4=x^2_{}$
$⇔5x-8+4\sqrt{x^2-5x+4}=x^2_{}$
$⇔4\sqrt{x^2-5x+4}=x^2-5x+8_{}$
$⇔16.(x^2-5x+4)=x^4+25x^2+64-10x^3+16x^2-80x_{}$
$⇔16x^2-80x+64=x^4+25x^2+64-10x^3+16x^2-80x_{}$
$⇔0=x^4+25x^2-10x^3_{}$
$⇔-x^4-25x^2+10x^3=0_{}$
$⇔-(x^4-10x^3+25x^2)=0_{}$
$⇔-x^2(x^2-10x+25)=0_{}$
$⇔-x^2(x-5)^2=0_{}$
$⇔x^2(x-5)^2=0_{}$
⇔ \(\left[ \begin{array}{l}x^2=0\\x-5=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=0(KTMĐK)\\x=5(TMĐK)\end{array} \right.\)
$Vậy_{}$ $x=5_{}$.
